The area of the region above the x-axis, bounded by the curve y= x, 0 x2, and the tangent …

Question

The area of the region above the $x$-axis, bounded by the curve $y=	an x$, $0\le x\le\dfrac{\pi}{2}$, and the tangent to the curve at $x=\dfrac{\pi}{4}$, is:

Accepted Answer

At $x=\dfrac\pi4$, $y=1$, and $\dfrac{dy}{dx}=\sec^2\dfrac\pi4=2$. Tangent: $y=2x+1-\dfrac\pi2$, meeting the $x$-axis at $x = \dfrac{\pi-2}{4}$.

$$\text{Area} = \int_0^{\pi/4}\tan x\,dx - \frac12\left(\frac\pi4-\frac{\pi-2}{4}\right)(1)$$

$$= [\ln\sec x]_0^{\pi/4} - \frac12\cdot\frac12 = \frac12\ln2 - \frac12$$

Solutions · 1